What is the equation of the normal line of $f(x)=(1-x)e^(3x)-e^x$ at $x=1$ ?

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Answer 1 · 2016-11-08T13:05:21

graph{((1-x)e^(3x)-e^x-y)(y+e-1/(e(e^2+1))(x-1))=0 [-3.844, 4.926, -3.28, 1.103]}

$y+e=1/(e(e^2+1))(x-1)$

The general equation for normal is

$y-f(x_0)=-1/f^'(x_0)(x-x_0)$

$x_0 = 1$

$f(x_0)=f(1)=-e$

$f^'(x)=-e^x(3xe^(2x)+1-2e^(2x))$

$f^'(x_0)=f^'(1)=-e(3e^2+1-2e^2)=-e(e^2+1)$

$y+e=1/(e(e^2+1))(x-1)$

What is the equation of the normal line of f(x)=(1-x)e^(3x)-e^xf(x)=(1-x)e^(3x)-e^x at x=1x=1?